Question

How do you simplify ${256^{ - \dfrac{7}{8}}}$?

Answer 1

Hint: First we know that the fractional exponents. I will show you what fractional exponents are: Suppose we are asked to simplify this:
\[{(625)^{\dfrac{1}{4}}}\]
This means that we have to find the ${4^{th}}$ root of $625$.
So in the form of a given below it will be like
\[{(625)^{\dfrac{1}{4}}} = \sqrt[4]{{625}} = 5\]
This is an example of fractional exponents.
We rewrite in the form of\[{a^{\dfrac{b}{c}}} = {\left( {{a^{\dfrac{1}{c}}}} \right)^b}\]
In the form of ${(a)^{ - b}}{(a)^b} = {a^{ - b + b}} = {a^0}$, we apply in the division.
Finally we get the fractional exponents value.

Complete step-by-step solution:
The given is ${256^{ - \dfrac{7}{8}}}$
This is the form of fractional exponents.
Let us consider $x = {256^{ - \dfrac{7}{8}}}$
First we separate the power values. We rewrite in the form of\[{a^{\dfrac{b}{c}}} = {\left( {{a^{\dfrac{1}{c}}}} \right)^b}\], hence we get
\[ = {\left( {{{\left( {256} \right)}^{\dfrac{1}{8}}}} \right)^{ - 7}}\]
Now we change the ${8^{th}}$ root, hence we get
$ = {\left( {\sqrt[8]{{256}}} \right)^{ - 7}}$
Now we have to find the${8^{th}}$root of $256$, hence we get
$ = {(2)^{ - 7}}$
Now multiply by ${(2)^7}$ in the numerator, hence we get
$ = {(2)^{ - 7}}{(2)^7}$
And divide by ${(2)^7}$, hence we get
$ = \dfrac{{{{(2)}^{ - 7}}{{(2)}^7}}}{{{{(2)}^7}}}$
Now in the form of ${(a)^{ - b}}{(a)^b} = {a^{ - b + b}} = {a^0}$, we apply in the division, hence we get
 $ = \dfrac{{{2^{ - 7}}^{ + 7}}}{{{{(2)}^7}}} = \dfrac{{{2^0}}}{{{2^7}}}$
The property is ${a^0} = 1$, so apply the numerator, hence we get
$ = \dfrac{1}{{{2^7}}}$
Now we expand the ${2^7}$ and we multiply by$2$in seven times, hence we get
$ = \dfrac{1}{{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}}$
$ = \dfrac{1}{{128}}$

The fractional exponent ${256^{ - \dfrac{7}{8}}}$ value is $\dfrac{1}{{128}}$

Note: One of the basic observations about integer exponents can be expressed as:
${a^p} \times {a^q} = {a^{p + q}}$
${a^p} \times {a^q} \times {a^r} = {a^{p + q + r}}$
And so on (post another question if it isn’t clear why this is true for integers)
If this observation is extended to fractional exponents we would have, for example, something like:
${a^{\dfrac{1}{3}}} \times {a^{\dfrac{1}{3}}} \times {a^{\dfrac{1}{3}}} = {a^{\dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}}} = {a^1} = a$
If three identical values multiplied together equal $a$ , then the values must be $\sqrt[3]{a}$